Now then, you will notice that the only difference between andro & test (or form & 4-OHT) is that the #17 carbonyl group is replaced by a hydroxyl group, thus activating the prohormone.
Converting a carbonyl group to hydroxyl is as easy as pie: Add a metalhydride (NaBH4 or LiAlH4) and water. Unfortunately, testosterone is not that easy.

Problem You'll notice that the lower left (A) ring has a carbonyl group in both cases, and this reaction is not selective on which oxo group it attacks.

In closing, I think harnessing the potential to convert prohormones to real hormones would be........awesome. If anything it should make for an interesting chemistry discussion, and who knows, maybe we'll learn something.

It is quite a bit more involved than what you have put forward. Converting a carbonyl (in this case a ketone) to a hydroxyl group is a reductive process. Simply adding acid in ether will do no such thing. Also, the lower left ring is called the 'A ring' in steriod chemistry. Reducing the A ring's ketone to a hydroxyl will not produce an estrogen. For a steroid to be classified as an estrogen, the A ring must be aromatic.

It's good that you've taken interest into the chemistry of steroids--keep studying and reading on the subject.

My bad on the ether reaction, I was going off memory and got mixed up with a grignard reaction. Whoops, wrong mechanism. I looked up the correct one, and edited my post.

Thanks for clarifying the makings of an estrogen, at the time of the posting, I wasn't sure if making the oxo group on the A ring would create estradiol or not (I wasn't paying attention to aromacity). My O-chem book doesn't spend as many pages on steroid chemistry as I would have liked (only 3 pages).

NaBH4 is not particularly effective as a reducing agent in water. There have been a number of cases in literature where the researchers noted that performing NaBH4 reduction in methanol is much more favorable. Ethanol is not as good as methanol, but it is still better than water. Additionally, adding a small amount of aqueous hydroxide ions slows the rate of borohydride decomposition. Since most steroids are water insoluble, the solvent of choice for reductive reactions is typically methanol.

LiAlH4 should never be dissolved in water unless one desires to form lithium hydroxide and aluminum hydroxide in an exotic fashion. Hydrogen gas will rapidly evolve with the danger of ignition. LiAlH4 reductions are most commonly carried out in a dry ether-type solvent such as diethyl ether or THF under an inert atmosphere.

Unfortunately, the molecule presented to be reduced has two ketones. The distinguishing factors, however are steric hinderance and the conjugated double bond next to the 3-one. Due to steric hinderance, the 3-one will normally react faster than the 17-one; however, since the 3-one is conjugated, it is more stable, and thus becomes less reactive. Unfortunately though, enones are suseptible to 1,4 reduction when NaBH4 is the reducing reagent. In fact, from various literature sources I've read, 1,4 reduction seems to be a preferrential route when using NaBH4. To avoid such side reaction problems, drastically lowering the temperature of the reaction mixture will provide the selectivity that we would want. Dry ice in acetone will provide -78 C which will cause the reaction to preferrentially reduce the 17-one to 17b-hydroxy. Of course, one would have to monitor this reaction via TLC to know when to stop the reaction.

-78 C seems to be the preferred temperature for many organic reactions. Thanks for all your help.

What do you do for a living klaus? You seem to have a very strong understanding in organic sythesis.